Morgan made a mistake when subtracting the rational expressions below 3t^2-4t+1/t+3-t^2+2t+2/t+3=2t^2-2t+3/t+3. What was Morgan's error? A. Morgan forgot to combine only li…

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asked Jul 12, 2023 140k views

1 Answer

Michael Ramos · Answer 1 · 2023-07-16T20:11:05+0000

Morgan forgot to distribute the negative sign to two of the terms in the second expression.

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Step-by-step explanation:

Focus on the numerators.

We have (3t^2-4t+1) as the first numerator and we subtract off (t^2+2t+2) as the second numerator.

Morgan needs to simplify (3t^2-4t+1)-(t^2+2t+2) for the numerator.

Mistakenly, she had these steps

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2+2t+2 .... her mistake made here

(3t^2-t^2)+(-4t+2t)+(1+2)

2t^2-2t+3

All of this applies to the numerator. The denominator stays at t+3 the entire time. So effectively we can ignore it on a temporary basis.

Here's what Morgan should have for her steps when simplifying the numerator.

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2-2t-2 ..... distribute the negative

(3t^2-t^2)+(-4t-2t)+(1-2)

2t^2-6t-1

Note in the second step, the negative outside flips the sign of each term in the second parenthesis.

Therefore,

$(3t^2-4t+1)/(t+3)-(t^2+2t+2)/(t+3)\\\\((3t^2-4t+1)-(t^2+2t+2))/(t+3)\\\\(3t^2-4t+1-t^2-2t-2)/(t+3)\\\\(2t^2-6t-1)/(t+3)\\\\$

which means
$(3t^2-4t+1)/(t+3)-(t^2+2t+2)/(t+3)=(2t^2-6t-1)/(t+3), \ \ \text{ where } t \\e -3\\\\$

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